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3 years ago

How much should Jordan budget monthly for an annual expenditure that coast $614.66? * TIMED!

Mathematics

2 answers:

poizon [28]3 years ago

6 0

Answer:

he should budget about $51.22

Step-by-step explanation:

annual mean one year since there is 12 months in a year you would want to divide 12 to $614.66

dlinn [17]3 years ago

3 0

Answer:

i think it's $51.22 Cause it said annual which means one year and it has 12 months so the actual price devide by 12

am not that sure tho cause the question is tricky

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Which of the following values would be an outlier in the following set? 1, 14, 17, 18, 19, 23

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1 would be the outlier

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For every four female students in a training class, there are two men. There are seven male students. What is the total of numbe

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There would be 14 female and 7 male, so the total number of students is 21.

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2 years ago

Between permanent life insurance and term life insurance, which typically has the lower premium and why?

adoni [48]

Term life insurance has the lower premium because term life insurance pays the face value to your beneficiary if you die within a certain set period of time, whereas whole life insurance covers you your entire life. (Option A)

<h3>What is insurance?</h3>

Insurance is when a third-party (the insurer) promises to indemnify another party (the insured) for losses they might suffer in the future in exchange for agreed upon payments.

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5 0

2 years ago

WSP CAN SOMEONE ACTUALLY HELP N EXPLAIN IM LOST

Alex787 [66]

Answer:

Option A

Step-by-step explanation:

Given expression:

3x² - 10x - 8

<u>Option A</u>

⇒ (x - 4)(3x + 2)

⇒ (x × 3x) + (2 × x) + (-4 × 3x) + (-4 × 2)

⇒ (3x²) + (2x) + (-12x) + (-8)

⇒ 3x² + 2x - 12x - 8

⇒ 3x² - 10x - 8

3x² - 10x - 8 = 3x² - 10x - 8 (Yes!)

<u>Option B</u>

⇒ (3x - 4)(x - 2)

⇒ (3x × x) + (3x × -2) + (-4 × x) + (-4 × -2)

⇒ (3x²) + (-6x) + (-4x) + (8)

⇒ 3x² - 6x - 4x + (8)

⇒ 3x² - 10x + 8

3x² - 10x - 8 = 3x² - 10x + 8 (No!)

<u>Option C</u>

⇒ (3x - 4)(x + 2)

⇒ (3x × x) + (3x × 2) + (-4 × x) + (-4 × 2)

⇒ (3x²) + (6x) + (-4x) + (-8)

⇒ 3x² + 6x - 4x - 8

⇒ 3x² + 2x - 8

3x² - 10x - 8 = 3x² + 2x - 8 (No!)

<u>Option D</u>

⇒ (3x - 2)(x - 4)

⇒ (3x × x) + (3x × -4) + (-2 × x) + (-2 × -4)

⇒ (3x²) + (-12x) + (-2x) + (8)

⇒ 3x² - 12x - 2x + 8

⇒ 3x² - 14x + 8

3x² - 10x - 8 = 3x² - 14x + 8 (No!)

Since the expression of option A has the same value as the given expression, option A is correct.

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2 years ago

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in exercises 15-20 find the vector component of u along a and the vecomponent of u orthogonal to a u=(2,1,1,2) a=(4,-4,2,-2)

Nimfa-mama [501]

Answer:

The component of $\vec u$ orthogonal to $\vec a$ is $\vec u_{\perp\,\vec a} = \left(\frac{9}{5},\frac{6}{5},\frac{9}{10},\frac{21}{10}\right)$ .

Step-by-step explanation:

Let $\vec u$ and $\vec a$ , from Linear Algebra we get that component of $\vec u$ parallel to $\vec a$ by using this formula:

$\vec u_{\parallel \,\vec a} = \frac{\vec u \bullet\vec a}{\|\vec a\|^{2}} \cdot \vec a$ (Eq. 1)

Where $\|\vec a\|$ is the norm of $\vec a$ , which is equal to $\|\vec a\| = \sqrt{\vec a\bullet \vec a}$ . (Eq. 2)

If we know that $\vec u =(2,1,1,2)$ and $\vec a=(4,-4,2,-2)$ , then we get that vector component of $\vec u$ parallel to $\vec a$ is:

$\vec u_{\parallel\,\vec a} = \left[\frac{(2)\cdot (4)+(1)\cdot (-4)+(1)\cdot (2)+(2)\cdot (-2)}{4^{2}+(-4)^{2}+2^{2}+(-2)^{2}} \right]\cdot (4,-4,2,-2)$

$\vec u_{\parallel\,\vec a} =\frac{1}{20}\cdot (4,-4,2,-2)$

$\vec u_{\parallel\,\vec a} =\left(\frac{1}{5},-\frac{1}{5},\frac{1}{10},-\frac{1}{10} \right)$

Lastly, we find the vector component of $\vec u$ orthogonal to $\vec a$ by applying this vector sum identity:

$\vec u_{\perp\,\vec a} = \vec u - \vec u_{\parallel\,\vec a}$ (Eq. 3)

If we get that $\vec u =(2,1,1,2)$ and $\vec u_{\parallel\,\vec a} =\left(\frac{1}{5},-\frac{1}{5},\frac{1}{10},-\frac{1}{10} \right)$ , the vector component of $\vec u$ is:

$\vec u_{\perp\,\vec a} = (2,1,1,2)-\left(\frac{1}{5},-\frac{1}{5},\frac{1}{10},-\frac{1}{10} \right)$

$\vec u_{\perp\,\vec a} = \left(\frac{9}{5},\frac{6}{5},\frac{9}{10},\frac{21}{10}\right)$

The component of $\vec u$ orthogonal to $\vec a$ is $\vec u_{\perp\,\vec a} = \left(\frac{9}{5},\frac{6}{5},\frac{9}{10},\frac{21}{10}\right)$ .

4 0

3 years ago